Consider the real-valued function f satisfying 2 f(sin x)+ f(cos x)=x. Then, the

domain of f(x) is R

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domain of f(x) is[-1, 1]

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range of f(x) is

(- \frac{2 π}{3}, \frac{π}{3}]

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range of f(x) is R

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Solution

The correct option is C range of f(x) is $(- \frac{2 π}{3}, \frac{π}{3}]$

$Given 2 f (sin x) + f (cos x) = x . . . . . [1] Replacing x by \frac{π}{2} - x, we get, 2 f (cos x) + f (sin x) = \frac{π}{2} - x . . . . . [2] Solving [1] and [2] by elimination method, we get, 3 f (sin x) = 3 x - \frac{π}{2} or, f (sin x) = x - \frac{π}{6} or, f (x) = {sin}^{- 1} x - \frac{π}{6} Clearly, f (x) has domain [- 1, 1] Also, {sin}^{- 1} x \in (- \frac{π}{2}, \frac{π}{2}] or, {sin}^{- 1} x - \frac{π}{6} \in (- \frac{2 π}{3}, \frac{π}{3}] \Rightarrow Range of f is (- \frac{2 π}{3}, \frac{π}{3}]$

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